Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{t + 3}{2t^2 + 14t + 24} \times \dfrac{t^2 + t - 20}{-t + 4} $
Solution: First factor out any common factors. $x = \dfrac{t + 3}{2(t^2 + 7t + 12)} \times \dfrac{t^2 + t - 20}{-(t - 4)} $ Then factor the quadratic expressions. $x = \dfrac {t + 3} {2(t + 3)(t + 4)} \times \dfrac {(t - 4)(t + 5)} {-(t - 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(t + 3) \times (t - 4)(t + 5) } { 2(t + 3)(t + 4) \times -(t - 4)} $ $x = \dfrac {(t - 4)(t + 5)(t + 3)} {-2(t + 3)(t + 4)(t - 4)} $ Notice that $(t + 3)$ and $(t - 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {(t - 4)(t + 5)\cancel{(t + 3)}} {-2\cancel{(t + 3)}(t + 4)(t - 4)} $ We are dividing by $t + 3$ , so $t + 3 \neq 0$ Therefore, $t \neq -3$ $x = \dfrac {\cancel{(t - 4)}(t + 5)\cancel{(t + 3)}} {-2\cancel{(t + 3)}(t + 4)\cancel{(t - 4)}} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $x = \dfrac {t + 5} {-2(t + 4)} $ $ x = \dfrac{-(t + 5)}{2(t + 4)}; t \neq -3; t \neq 4 $